- #1
stephenklein
- 6
- 3
- Homework Statement
- A circular disk of radius R rotates with angular velocity (##\Omega##). The center of the disk is
at rest in an inertial frame. Light of frequency (##\omega##) is radiated from a point source at the center of the disk. What is the frequency (##\omega '##) measured by an observer attached to the rim of the rotating disk? Solve this problem in three different ways:
(a) Use the fact that if a particle has four-momentum ##\bar{p} = (E, \vec p)## in an inertial frame O then the energy (E ') seen by an observer with four-velocity ##U_{obs}## is ##E' = -\bar{p} \cdot \bar{e_{t'}}##.
(b) Use the Lorentz transformation of components of the photon four-momentum.
(c) In the frame of the rotating disk the effect of the gravitational field is to blue-shift
the photon. (*Use the results of (c) in the Rotating Space Station probe question).
*The referenced question was to derive the proper time of an observer in the hull of a rotating space station, the answer given as ##\tau = \frac{t'}{\gamma (v)}##, where the relative velocity v is ##r\Omega##, the radial distance from the space station's axis of rotation times a constant angular speed ##\Omega##.
- Relevant Equations
- Energy of a particle according to any inertial observer:
$$E' = -\bar{p} \cdot \bar{U_{obs}}$$
Lorentz Transformation as a partial derivative:
$$\Lambda^{\alpha '}_{\alpha} = \frac{\partial x^{\alpha '}}{\partial x^{\alpha}} $$
Proper time for observer on the rim of the disk:
$$\tau = \frac{t'}{\gamma (v)} $$
**I realize some of my inline math delimiters '\(' and '\)' are not acting on the text for some reason, and it looks clunky. I spend 20-30 minutes trying to understand why this is, but I can't. My limited LaTeX experience is in Overleaf, and these delimiters work fine in that compiler. My apologies for the clunky format**
**Edit: Big thanks to Mark44 for the LaTeX help**
Part a:
Scalar products must happen using the same coordinate basis for each vector, so I should find ##\bar{e_{t'}}##, the rotating observer's four-velocity, in the coordinate basis of the center of the disk frame, which is the coordinate basis of the photon's four-momentum. In cylindrical coordinates, the frames are related by the following (these relations were given in the previous question on the assignment):
$$t'=t$$
$$r'=r$$
$$\phi ' = \phi - \omega t$$
$$z'=z$$
Apply the Lorentz transformation to the observer's four-velocity, ##\bar{e_{0'}}##, to get its components in the inertial frame:
$$\bar{U_{obs}}=\bar{e_{0'}}=\frac{\partial x^{\alpha}}{\partial x^{0'}} \bar{e_{\alpha}}$$
$$\bar{U_{obs}}= \begin{pmatrix} 1 \\ 0 \\ \Omega\\ 0 \end{pmatrix}$$
Four-momentum, using natural units so c=1, is:
\begin{pmatrix} h\omega \\ h\omega \\ 0 \\ 0 \end{pmatrix}
From this, it's clear that the (negative) scalar product will just give me back the original energy, ##h\omega##, which I know is wrong. I'm guessing I'm going wrong with the Lorentz transformation, but that's exactly how the transformation works according to my notes.
Part b:
Well, the Lorentz transformation for the photon's four-momentum should be the inverse of the transformation I tried in part a. So, in the primed frame (the rotating one):
$$p^{\alpha '} = \frac{\partial x^{\alpha '}}{\partial x^{\alpha}} p^{\alpha}$$
$$p^{\alpha '} = \begin{pmatrix} 1 & 0 & -\Omega & 0 \\ 0 & 1 & 0 & 0 \\ \frac{1}{\Omega} & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} h\omega \\ h\omega \\ 0 \\ 0 \end{pmatrix}$$
$$p^{\alpha '} = \begin{pmatrix} h\omega \\ h\omega \\ \frac{h\omega}{\Omega} \\ 0 \end{pmatrix}$$
Well, clearly something is wrong, because this is not a null vector. The components for the Lorentz matrix were obtained by applying partial derivatives to the coordinate relations, and substituting t=t' when finding the ##\Lambda^{2}_{0}## component:
$$t = \frac{\phi - \phi '}{\Omega}$$
$$t'= \frac{\phi - \phi '}{\Omega}$$
This answer, along with my answer for part a, seems to indicate I must be misunderstanding something regarding the transformations.
Part c:
Proper time in the rotating frame:
$$\tau = \frac{t}{\gamma}$$
So the frequency should be ##\frac{1}{\tau}##:
$$\omega ' = \frac{1}{\tau} = \frac{\gamma}{t}$$
Here, ##\gamma## is ##\sqrt{1-(\frac{r\Omega}{c})^{2}}##, and ##\frac{1}{t}## is the photon's frequency in the inertial frame. So, ##\omega '= \omega \sqrt{1-(\frac{r\Omega}{c})^{2}}##. I'm fairly confident this answer is actually right.
**Edit: Big thanks to Mark44 for the LaTeX help**
Part a:
Scalar products must happen using the same coordinate basis for each vector, so I should find ##\bar{e_{t'}}##, the rotating observer's four-velocity, in the coordinate basis of the center of the disk frame, which is the coordinate basis of the photon's four-momentum. In cylindrical coordinates, the frames are related by the following (these relations were given in the previous question on the assignment):
$$t'=t$$
$$r'=r$$
$$\phi ' = \phi - \omega t$$
$$z'=z$$
Apply the Lorentz transformation to the observer's four-velocity, ##\bar{e_{0'}}##, to get its components in the inertial frame:
$$\bar{U_{obs}}=\bar{e_{0'}}=\frac{\partial x^{\alpha}}{\partial x^{0'}} \bar{e_{\alpha}}$$
$$\bar{U_{obs}}= \begin{pmatrix} 1 \\ 0 \\ \Omega\\ 0 \end{pmatrix}$$
Four-momentum, using natural units so c=1, is:
\begin{pmatrix} h\omega \\ h\omega \\ 0 \\ 0 \end{pmatrix}
From this, it's clear that the (negative) scalar product will just give me back the original energy, ##h\omega##, which I know is wrong. I'm guessing I'm going wrong with the Lorentz transformation, but that's exactly how the transformation works according to my notes.
Part b:
Well, the Lorentz transformation for the photon's four-momentum should be the inverse of the transformation I tried in part a. So, in the primed frame (the rotating one):
$$p^{\alpha '} = \frac{\partial x^{\alpha '}}{\partial x^{\alpha}} p^{\alpha}$$
$$p^{\alpha '} = \begin{pmatrix} 1 & 0 & -\Omega & 0 \\ 0 & 1 & 0 & 0 \\ \frac{1}{\Omega} & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} h\omega \\ h\omega \\ 0 \\ 0 \end{pmatrix}$$
$$p^{\alpha '} = \begin{pmatrix} h\omega \\ h\omega \\ \frac{h\omega}{\Omega} \\ 0 \end{pmatrix}$$
Well, clearly something is wrong, because this is not a null vector. The components for the Lorentz matrix were obtained by applying partial derivatives to the coordinate relations, and substituting t=t' when finding the ##\Lambda^{2}_{0}## component:
$$t = \frac{\phi - \phi '}{\Omega}$$
$$t'= \frac{\phi - \phi '}{\Omega}$$
This answer, along with my answer for part a, seems to indicate I must be misunderstanding something regarding the transformations.
Part c:
Proper time in the rotating frame:
$$\tau = \frac{t}{\gamma}$$
So the frequency should be ##\frac{1}{\tau}##:
$$\omega ' = \frac{1}{\tau} = \frac{\gamma}{t}$$
Here, ##\gamma## is ##\sqrt{1-(\frac{r\Omega}{c})^{2}}##, and ##\frac{1}{t}## is the photon's frequency in the inertial frame. So, ##\omega '= \omega \sqrt{1-(\frac{r\Omega}{c})^{2}}##. I'm fairly confident this answer is actually right.
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